Question: Evaluate the iterated integral. $ \int_1^{e^2} \left( \int_{-2}^2 \dfrac{x + y}{x^2} \, dy \right) dx =$ Choose 1 answer: Choose 1 answer: (Choice A) A $e^2 - 4$ (Choice B) B $\dfrac{e^4 + 4e^2}{e^2 - 1}$ (Choice C) C $8$ (Choice D) D $\dfrac{2e^4 - 2}{e^2}$
Explanation: Evaluate the inner integral: $\begin{aligned} \int_1^{e^2} \left( \int_{-2}^2 \dfrac{x + y}{x^2} \, dy \right) dx &= \int_1^{e^2} \left( \int_{-2}^2 \dfrac{1}{x} + \dfrac{y}{x^2} \, dy \right) dx \\ \\ &= \int_1^{e^2} \left[ \dfrac{y}{x} + \dfrac{y^2}{2x^2} \right]_{-2}^2 dx \\ \\ &= \int_1^{e^2} \dfrac{2}{x} + \dfrac{4}{2x^2} - \dfrac{-2}{x} - \dfrac{4}{2x^2} \, dx \\ \\ &= \int_1^{e^2} \dfrac{4}{x} \, dx \end{aligned}$ Evaluate the outer integral: $\begin{aligned} \int_1^{e^2} \dfrac{4}{x} \, dx &= 4 \ln(x) \bigg|_1^{e^2} \\ \\ &= 4 \ln \left( e^2 \right) - 4 \ln(1) \\ \\ &= 8 \end{aligned}$ The answer: $ \int_1^{e^2} \left( \int_{-2}^2 \dfrac{x + y}{x^2} \, dy \right) dx = 8$